The original circlex2+y2−6x−10y+30=0has centreC=(3,5), radius r=32+52−30=4=2.We reflect the centre across the line2x+y−2=0Reflection formula gives the image point C′(x′,y′) :d=22+122(3)+1(5)−2=59.Thus,x′=3−2(2)(59)=3−536=−521y′=5−2(1)(59)=5−518=57So the image centre isC′=(−521,57).Radius remains 2 .Its equation is:(x−x′)2+(y−y′)2=4. Expanding:x2+y2+542x−514y+(x′2+y′2−4)=0Compute constant:x′2+y′2=25441+2549=25490=598So:γ=598−4=598−20=578.Thus:α+β+γ=542−514+578=5106=21.2.{[α+β+γ]=⌊21.2⌋=21.}